Question: Is the function given below continuous/differentiable at $x=4$ ? $f(x)=\begin{cases} 2x+3&,&x\leq4 \\\\ x^2-5&,&x>4 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Checking for continuity at $x=4$ For the function to be continuous at $x=4$, we need the two-sided limit $\lim_{x\to 4}f(x)$ to exist and be equal to $f(4)$. This is the same as requiring that the two one-sided limits $\lim_{x\to 4^-}f(x)$ and $\lim_{x\to 4^+}f(x)$ exist and are equal to $f(4)$. According to $f$ 's definition, $f(4)=2(4)+3=11$. $\lim_{x\to 4^-}f(x)$ $2x+3$ evaluated at $x=4$ is equal to $11$. Since $2x+3$ is continuous, we can be certain that $\lim_{x\to 4^-}f(x)=11$. $\lim_{x\to 4^+}f(x)$ $x^2-5$ evaluated at $x=4$ is equal to $11$. Since $x^2-5$ is continuous, we can be certain that $\lim_{x\to 4^+}f(x)=11$. We saw that the two one-sided limits exist and are equal to $f(4)$, so the function is continuous at $x=4$. Checking for differentiability at $x=4$ For the function to be differentiable at $x=4$, we need the two-sided limit $\lim_{x\to 4}\dfrac{f(x)-f(4)}{x-4}=\lim_{x\to 4}\dfrac{f(x)-11}{x-4}$ to exist. This is the same as requiring that the two one-sided limits $\lim_{x\to 4^-}\dfrac{f(x)-11}{x-4}$ and $\lim_{x\to 4^+}\dfrac{f(x)-11}{x-4}$ exist and have the same value. $\lim_{x\to 4^-}\dfrac{f(x)-11}{x-4}=2$ $\lim_{x\to 4^+}\dfrac{f(x)-11}{x-4}=8$ The two limits exist, but they are not equal. Therefore, the function is not differentiable at $x=4$. Graphically, the function has a sharp turn at this point. [I would like to see that, please!] In conclusion, the function is continuous but not differentiable at $x=4$.